Normalized equation for particle in a ring

[Titan97]

Homework Statement

Normalized equation for particle in a ring, where V=0 on a ring of radius 'a' and infinite everywhere else.

Homework Equations

The Attempt at a Solution

Replcing x by rθ,
$$-\frac{\hbar^2}{2I}\frac{\partial^2\psi}{\partial\theta^2}=E\psi$$
By guess, I found out that $Ae^{ik\theta}$ is an eigenfunction of the hamiltonian.
To find $A$, I used:
$$\int_0^{2\pi}|\psi|^2rd\theta=1$$
But in the solution, they used
$$\int_0^{2\pi}|\psi|^2d\theta=1$$So the normalization constants were different. Which one is correct?

 

[PeroK]

Homework Statement

Normalized equation for particle in a ring, where V=0 on a ring of radius 'a' and infinite everywhere else.

Homework Equations

The Attempt at a Solution

Replcing x by rθ,
$$-\frac{\hbar^2}{2I}\frac{\partial^2\psi}{\partial\theta^2}=E\psi$$
By guess, I found out that $Ae^{ik\theta}$ is an eigenfunction of the hamiltonian.
To find $A$, I used:
$$\int_0^{2\pi}|\psi|^2rd\theta=1$$
But in the solution, they used
$$\int_0^{2\pi}|\psi|^2d\theta=1$$So the normalization constants were different. Which one is correct?

You are not integrating a function of $r, \theta$ over two dimensions. It's a simple one-dimensional integral for $\theta$.

 

[Titan97]

the particle moves along a ring. so the elemental length should be rdtheta

 

[PeroK]

the particle moves along a ring. so the elemental length should be rdtheta

You've defined your $\psi$ in terms of $\theta$. $|\psi|^2$ is the probability density function in terms of $\theta$, not in terms of $l$ where $l$ is the arc length along the ring.

Normalisation means that the total proability is 1, so the simple integral of $|\psi(\theta)|^2 d\theta$ is required.

If you transformed $\psi$ into a function of arc length, $l$, then you would integrate $|\psi(l)|^2 dl$.

To illustrate this, a constant pdf for $\theta$ would be $p(\theta) = \frac{1}{2 \pi}$, whereas a constant pdf for $l$ would be $p(l) = \frac{1}{2 \pi a}$